Q.
Three identical charges, each of $2\mu C$ are placed at the vertices of a triangle ABC as shown in the figure
If $AB + AC = 12\ cm$ and $AB.AC = 32\ cm^{2},$ the potential energy of the charge at $A$ is
A) $1.53\ J$
B) $5.31\ J$
C) $3.15\ J$
D) $1.35\ J$
Explanation
(d) $AB + AC = 12\ cm$ …(i)
$AB.AC = 32\ cm^{2}$
$\therefore AB - AC = \sqrt{(AB + AC)^{2} - 4\ AB.AC}$
$AB - AC = 4$ …(ii)
From equation (i) and (ii)
$AB = 8\ cm;AC = 4\ cm$
Potential energy at point $A$
$V_{A} = \frac{1}{4\pi\varepsilon_{0}}q_{1}q_{2}\left\lbrack \frac{1}{AB} + \frac{1}{AC} \right\rbrack$
$= \frac{9 \times 10^{9} \times 2 \times 2 \times 10^{- 12}}{10^{- 2}}\left( \frac{1}{8} + \frac{1}{4} \right) = 1.35\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.